3.12.55 \(\int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx\) [1155]

3.12.55.1 Optimal result
3.12.55.2 Mathematica [A] (verified)
3.12.55.3 Rubi [A] (verified)
3.12.55.4 Maple [B] (verified)
3.12.55.5 Fricas [B] (verification not implemented)
3.12.55.6 Sympy [F]
3.12.55.7 Maxima [F(-2)]
3.12.55.8 Giac [F(-2)]
3.12.55.9 Mupad [F(-1)]

3.12.55.1 Optimal result

Integrand size = 32, antiderivative size = 200 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {\sqrt [4]{-1} a^{5/2} (c+5 i d) \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {c-i d} f}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f} \]

output
-(-1)^(1/4)*a^(5/2)*(c+5*I*d)*arctanh((-1)^(3/4)*d^(1/2)*(a+I*a*tan(f*x+e) 
)^(1/2)/a^(1/2)/(c+d*tan(f*x+e))^(1/2))/d^(3/2)/f-4*I*a^(5/2)*arctanh(2^(1 
/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a*tan(f*x+e))^(1/2)) 
*2^(1/2)/f/(c-I*d)^(1/2)-a^2*(a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/ 
2)/d/f
 
3.12.55.2 Mathematica [A] (verified)

Time = 6.13 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.28 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {4 i \sqrt {2} a^2 \sqrt {-a c+i a d} \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{c f-i d f}+\frac {-a^2 \sqrt {d} \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))-(-1)^{3/4} a^{5/2} \sqrt {c+i d} (c+5 i d) \arcsin \left (\frac {\sqrt [4]{-1} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+i d}}\right ) \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}}{d^{3/2} f \sqrt {c+d \tan (e+f x)}} \]

input
Integrate[(a + I*a*Tan[e + f*x])^(5/2)/Sqrt[c + d*Tan[e + f*x]],x]
 
output
((4*I)*Sqrt[2]*a^2*Sqrt[-(a*c) + I*a*d]*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[ 
a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])])/(c*f - I*d*f 
) + (-(a^2*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])) - (-1) 
^(3/4)*a^(5/2)*Sqrt[c + I*d]*(c + (5*I)*d)*ArcSin[((-1)^(1/4)*Sqrt[d]*Sqrt 
[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + I*d])]*Sqrt[(c + d*Tan[e + f*x]) 
/(c + I*d)])/(d^(3/2)*f*Sqrt[c + d*Tan[e + f*x]])
 
3.12.55.3 Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.344, Rules used = {3042, 4039, 27, 3042, 4084, 3042, 4027, 221, 4082, 66, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}}dx\)

\(\Big \downarrow \) 4039

\(\displaystyle \frac {a \int \frac {\sqrt {i \tan (e+f x) a+a} (a (i c+3 d)+a (c+5 i d) \tan (e+f x))}{2 \sqrt {c+d \tan (e+f x)}}dx}{d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \int \frac {\sqrt {i \tan (e+f x) a+a} (a (i c+3 d)+a (c+5 i d) \tan (e+f x))}{\sqrt {c+d \tan (e+f x)}}dx}{2 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \int \frac {\sqrt {i \tan (e+f x) a+a} (a (i c+3 d)+a (c+5 i d) \tan (e+f x))}{\sqrt {c+d \tan (e+f x)}}dx}{2 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\)

\(\Big \downarrow \) 4084

\(\displaystyle \frac {a \left (8 a d \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx+(-5 d+i c) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx\right )}{2 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (8 a d \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx+(-5 d+i c) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx\right )}{2 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {a \left ((-5 d+i c) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx-\frac {16 i a^3 d \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f}\right )}{2 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {a \left ((-5 d+i c) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx-\frac {8 i \sqrt {2} a^{3/2} d \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f \sqrt {c-i d}}\right )}{2 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {a \left (\frac {a^2 (-5 d+i c) \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{f}-\frac {8 i \sqrt {2} a^{3/2} d \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f \sqrt {c-i d}}\right )}{2 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {a \left (\frac {2 a^2 (-5 d+i c) \int \frac {1}{i a-\frac {d (i \tan (e+f x) a+a)}{c+d \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}}{f}-\frac {8 i \sqrt {2} a^{3/2} d \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f \sqrt {c-i d}}\right )}{2 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {a \left (\frac {2 (-1)^{3/4} a^{3/2} (-5 d+i c) \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f}-\frac {8 i \sqrt {2} a^{3/2} d \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f \sqrt {c-i d}}\right )}{2 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\)

input
Int[(a + I*a*Tan[e + f*x])^(5/2)/Sqrt[c + d*Tan[e + f*x]],x]
 
output
(a*((2*(-1)^(3/4)*a^(3/2)*(I*c - 5*d)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + 
 I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[d]*f) - ((8 
*I)*Sqrt[2]*a^(3/2)*d*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/( 
Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[c - I*d]*f)))/(2*d) - (a 
^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(d*f)
 

3.12.55.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 66
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4027
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f)   Subst[Int[1/(a*c - b*d - 2* 
a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; 
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N 
eQ[c^2 + d^2, 0]
 

rule 4039
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + 
 d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) 
Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + 
a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x 
] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] 
 && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4082
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(B/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], 
x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
 

rule 4084
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*b + a*B)/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] 
 - Simp[B/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ 
e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - 
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
 
3.12.55.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1294 vs. \(2 (157 ) = 314\).

Time = 1.30 (sec) , antiderivative size = 1295, normalized size of antiderivative = 6.48

method result size
derivativedivides \(\text {Expression too large to display}\) \(1295\)
default \(\text {Expression too large to display}\) \(1295\)

input
int((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOS 
E)
 
output
1/4/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a^2*(4*I*2^(1/2)*( 
-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*( 
c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d+I*2^(1/2)*( 
-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*( 
c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^3-4*I*d^2*ln( 
1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2 
)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*2^(1/2)*(-a*(I*d-c))^(1/2)+4*I*ln((3 
*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)* 
(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2) 
*a*c*d+I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a* 
(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2)) 
*a*c*d^2+4*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*( 
-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e) 
+I))*(I*a*d)^(1/2)*a*d^2+3*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan( 
f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)* 
(-a*(I*d-c))^(1/2)*a*c^2*d+3*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*ta 
n(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2 
)*(-a*(I*d-c))^(1/2)*a*d^3-2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*( 
I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^2+4*ln(1/2*(2*I*a*d*tan(f*x+e)+I 
*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(...
 
3.12.55.5 Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 781 vs. \(2 (150) = 300\).

Time = 0.26 (sec) , antiderivative size = 781, normalized size of antiderivative = 3.90 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {2 \, \sqrt {2} a^{2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} + d f \sqrt {\frac {i \, a^{5} c^{2} - 10 \, a^{5} c d - 25 i \, a^{5} d^{2}}{d^{3} f^{2}}} \log \left (\frac {{\left (2 \, d^{2} f \sqrt {\frac {i \, a^{5} c^{2} - 10 \, a^{5} c d - 25 i \, a^{5} d^{2}}{d^{3} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left (-i \, a^{2} c + 5 \, a^{2} d + {\left (-i \, a^{2} c + 5 \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{-i \, a^{2} c + 5 \, a^{2} d}\right ) - d f \sqrt {\frac {i \, a^{5} c^{2} - 10 \, a^{5} c d - 25 i \, a^{5} d^{2}}{d^{3} f^{2}}} \log \left (-\frac {{\left (2 \, d^{2} f \sqrt {\frac {i \, a^{5} c^{2} - 10 \, a^{5} c d - 25 i \, a^{5} d^{2}}{d^{3} f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left (-i \, a^{2} c + 5 \, a^{2} d + {\left (-i \, a^{2} c + 5 \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{-i \, a^{2} c + 5 \, a^{2} d}\right ) - \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c + d\right )} f^{2}}} d f \log \left (\frac {{\left (\sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c + d\right )} f^{2}}} {\left (i \, c + d\right )} f e^{\left (i \, f x + i \, e\right )} + 4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2}}\right ) + \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c + d\right )} f^{2}}} d f \log \left (\frac {{\left (\sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c + d\right )} f^{2}}} {\left (-i \, c - d\right )} f e^{\left (i \, f x + i \, e\right )} + 4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2}}\right )}{2 \, d f} \]

input
integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="fr 
icas")
 
output
-1/2*(2*sqrt(2)*a^2*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I 
*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) + d* 
f*sqrt((I*a^5*c^2 - 10*a^5*c*d - 25*I*a^5*d^2)/(d^3*f^2))*log((2*d^2*f*sqr 
t((I*a^5*c^2 - 10*a^5*c*d - 25*I*a^5*d^2)/(d^3*f^2))*e^(I*f*x + I*e) + sqr 
t(2)*(-I*a^2*c + 5*a^2*d + (-I*a^2*c + 5*a^2*d)*e^(2*I*f*x + 2*I*e))*sqrt( 
((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt( 
a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(-I*a^2*c + 5*a^2*d)) - d*f 
*sqrt((I*a^5*c^2 - 10*a^5*c*d - 25*I*a^5*d^2)/(d^3*f^2))*log(-(2*d^2*f*sqr 
t((I*a^5*c^2 - 10*a^5*c*d - 25*I*a^5*d^2)/(d^3*f^2))*e^(I*f*x + I*e) - sqr 
t(2)*(-I*a^2*c + 5*a^2*d + (-I*a^2*c + 5*a^2*d)*e^(2*I*f*x + 2*I*e))*sqrt( 
((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt( 
a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(-I*a^2*c + 5*a^2*d)) - sqr 
t(-32*I*a^5/((I*c + d)*f^2))*d*f*log(1/4*(sqrt(-32*I*a^5/((I*c + d)*f^2))* 
(I*c + d)*f*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sq 
rt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sq 
rt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2) + sqrt(-32*I*a^5/(( 
I*c + d)*f^2))*d*f*log(1/4*(sqrt(-32*I*a^5/((I*c + d)*f^2))*(-I*c - d)*f*e 
^(I*f*x + I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sqrt(((c - I*d) 
*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I* 
f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2))/(d*f)
 
3.12.55.6 Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]

input
integrate((a+I*a*tan(f*x+e))**(5/2)/(c+d*tan(f*x+e))**(1/2),x)
 
output
Integral((I*a*(tan(e + f*x) - I))**(5/2)/sqrt(c + d*tan(e + f*x)), x)
 
3.12.55.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Exception raised: ValueError} \]

input
integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="ma 
xima")
 
output
Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume((d^2-2*c*d-c^2)>0)', see `assume 
?` for mor
 
3.12.55.8 Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \]

input
integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="gi 
ac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Non regular value [0,0] was discard 
ed and replaced randomly by 0=[11,77]Warning, replacing 11 by 3, a substit 
ution var
 
3.12.55.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]

input
int((a + a*tan(e + f*x)*1i)^(5/2)/(c + d*tan(e + f*x))^(1/2),x)
 
output
int((a + a*tan(e + f*x)*1i)^(5/2)/(c + d*tan(e + f*x))^(1/2), x)